2. Because OA is a transversal crossing the parallels A'A and OR, it is in the same plane with them. Also, because OR 1 plane MN, OR 1 OA' in that plane (§ 584). The same things are true of PB', PB, and PS.

3. Because A'OR and B'OB are right angles, AOA' is the complement of the angle AOR, and BPB' is the complement of the equal angle BPS. Comparing with (1),

611. Between two lines not in the same plane one, and only one, common perpendicular can be drawn. Hypothesis. AB, CD, two lines not in the same plane, and therefore passing each other without intersecting.

Conclusion. There is one line, and no more, perpendicular to both AB and CD.

Proof. I. Through one line, say CD, pass a plane, and let it turn round on CD until it is parallel to AB. Let MN be this plane. Let A'B' be the projection of AB on the plane MN, and let O be the point in which this projection. intersects CD. Then

1. Every point of A'B' is fixed by dropping a perpendicular from some point of AB. Let P be the point, of which O is the projection. Then PO 1 plane MN (§ 597).

2. Because PO is perpendicular to MN, it is perpendicular to both the lines A'B' and CD, which lie in MN.

3. Because PO is perpendicular to A'B', it is also perpendicular to AB, which is parallel to A'B' (§ 72).

Therefore OP is perpendicular to both the lines AB and CD. Q.E.D.

II. If there is any other common perpendicular, let it be P'Q. Through Q draw, in the plane MN, QR || AB. Then— 4. Because P'Q is perpendicular to AB, it is also perpendicular to QR, which is parallel to AB.

5. Because P'Q is perpendicular to both the lines QR and CD, it is perpendicular to their plane MN.

6. But, because A'B' is the projection of AB, the foot of the perpendicular from P' on the plane must fall on some point of A'B'. Let O' be this point.

Therefore from the point P' are dropped two perpendiculars P'O' and P'Q upon plane MN, which is impossible (§ 585). Therefore P'Q is not a common perpendicular to the lines AB and CD, and PO is the only common perpendicular. Q.E.D.

THEOREM XVII.

612. The least distance between two indefinite lines which do not intersect each other is their common perpendicular.

Hypothesis. a, b, two lines in space, the one being supposed to lie behind the other, so that they do not intersect. Conclusion. No line which

is not perpendicular to both lines can be the shortest line between them.

R

Proof. If possible, suppose that some line PQ which does not make a right angle with a is the shortest line. From P drop a perpendicular PR upon a.

PR < PQ.

Therefore PQ is not the shortest line.

Then

But since the lines do not intersect, some line must be the shortest.

Therefore this line is one making a right angle with both lines. Q.E.D.

CHAPTER II.

RELATIONS OF TWO OR MORE PLANES.

THEOREM XVIII.

613. Two planes are parallel if any two intersecting lines on the one are both parallel to the other plane. Hypothesis. AB, CD, two intersecting lines lying in the

plane MN, and each of them

parallel to the plane PQ.

Conclusion. The planes MN and PQ are parallel.

Proof. 1. If the planes are not parallel, they must intersect in a straight line. lying in both planes, and therefore in the plane MN. Let us call this line X.

2. Because the lines AB and CD are not parallel, the line X must intersect at least one of them.

3. Because the line X is also in the plane PQ, the line AB or CD, where it intersects X, would also intersect the plane PQ.

4. But, by hypothesis, these lines are parallel to the plane PQ, and therefore cannot meet it.

5. Therefore there can be no such line as X, and the plane MN can nowhere meet the plane PQ. Therefore these planes are parallel. Q.E.D.

THEOREM XIX.

614. The lines in which two parallel planes intersect a third plane are parallel.

Let us call A and B the parallel planes,* and X the third or intersecting plane. Then

1. The lines of intersection are in one plane, because they both lie in the plane X

2. Because one of the lines of intersection lies in the plane A, and the other in the plane B, parallel to it, and because these planes never meet, the lines can never meet.

Therefore the lines are in one plane and can never meet, and so are parallel, by definition. Q.E.D.

THEOREM XX.

615. Parallel planes intercept equal segments of parallel lines.

Hypothesis. MN, PQ, two parallel planes; AB, CD, two parallel lines intersecting the planes in the points A, B,

2. But because the lines AC and BD also lie in the respective planes MN and PQ, they are the lines of intersection of these planes with the plane ABCD.

3. Because the planes MN and PQ are parallel (hyp.), the lines of intersection AC and BD are parallel (§ 614).

4. Because AB || CD (hyp.) and AC || BD, as just shown, ABCD is a parallelogram. Therefore

*This theorem is so simple that the student can imagine the figure which is to embody the hypothesis and conclusion better than it can be represented in a diagram. We therefore give the demonstration without a diagram.

THEOREM XXI.

616. Planes perpendicular to the same straight line are parallel or coincident.

Hypothesis. Two planes, MN and PQ; OR, a line perpendicular to each of these planes. Conclusion. The planes are

parallel.

Proof. If they are not parallel, they must intersect. If they intersect, call X any point on the line of intersection and join OX, RX. Then

1. Because OX is in the plane

M

MN, it is perpendicular to OR, a perpendicular line to the plane.

2. Because RX is in the plane PQ, it is also perpendicular to OR.

3. Therefore from the same point, X, we have two perpendiculars, XO and XR, to the same straight line, OR, which is impossible.

4. Therefore the planes never meet, and so are parallel.

Q.E.D. 617. Corollary. Conversely, a straight line perpendicular to a plane is also perpendicular to every parallel plane.

THEOREM XXII.

618. A straight line makes equal angles with parallel planes.

Hypothesis. MN, PQ, two parallel planes; AB, a straight line intersecting these planes at the points E and F; EA', FA", the projections of this line upon the respective planes.

P

M

E

B